Answer
$$\sin\Big(\frac{3\pi}{2}-x\Big)=-\cos x$$
Work Step by Step
We know that $$\frac{3\pi}{2}=\pi+\frac{\pi}{2}$$
Therefore,
$$\sin\frac{3\pi}{2}=\sin\Big(\pi+\frac{\pi}{2}\Big)=\sin\pi\cos\frac{\pi}{2}+\cos\pi\sin\frac{\pi}{2}=0\times0+(-1)\times1=-1$$
$$\cos\frac{3\pi}{2}=\cos\Big(\pi+\frac{\pi}{2}\Big)=\cos\pi\cos\frac{\pi}{2}-\sin\pi\sin\frac{\pi}{2}=(-1)\times0+0\times1=0$$
Now return to the given quantity and apply the addition formula for sine:
$$\sin\Big(\frac{3\pi}{2}-x\Big)=\sin\frac{3\pi}{2}\cos x-\cos\frac{3\pi}{2}\sin x$$
$$\sin\Big(\frac{3\pi}{2}-x\Big)=(-1)\times\cos x-0\times\sin x$$
$$\sin\Big(\frac{3\pi}{2}-x\Big)=-\cos x$$
